Semisimplicity of Frobenius on *integral* Tate module Let $K$ be a number field and $A/K$ an Abelian variety; let $l$ be a (rational) prime.  Do there exist infinitely many primes $\mathfrak{p}$ of $K$ such that the Frobenius at $\mathfrak{p}$ acts semisimply on the integral Tate module $$T_l(A)\otimes \overline{\mathbb{Z}_\ell}?$$
By this I mean that there is a $\overline{\mathbb{Z}_\ell}$-basis of eigenvectors for the Frobenius action.
The answer is well-known to be yes if the integral Tate module is replaced with the rational Tate module (this goes back to Weil).  If this is true, it should probably boil down to some Chebotarev argument, but I'm blanking at the moment.
Some remarks:  As znt notes, the answer is "yes" if the image of the representation $G_K\to GL(A[\ell])$ contains a matrix with distinct eigenvalues, by Chebotarev.  
Moreover, one can show that for any prime $\mathfrak{p}$ at which $A$ has good reduction, the Frobenius at $\mathfrak{p}$ acts semisimply on the $\ell$-adic Tate module for almost all $\ell$ (this follows from the theory of Frobenius tori).  If the Mumford-Tate conjecture holds for $A$, I think it's enough to find any torus with good reduction at $\ell$ in the Mumford-Tate group of $A$, though I haven't written out the details.  In any case, to find a counterexample it should be enough to find an Abelian variety over $K$ whose Mumford-Tate group contains no torus with good reduction at $\ell$.
One may also show the result holds if the image of the map $G_k\to GL(T_\ell(A))$ contains any element with distinct eigenvalues which is semi-simple over $\overline{\mathbb{Z}_\ell}$.
 A: Probably not.
Let $E_1$ and $E_2$ be two distinct elliptic curves with identical $\ell$-torsion representations, and let $A$ be $E_1 \times E_2$ mod the diagonal $\ell$-torsion representation.
Then at any prime $\mathfrak p$ where Frobenius acts semisimply on $A[\ell]$, $E_1$ and $E_2$ are isogenous. Indeed, if the eigenvalues of Frobenius of $E_1$ and $E_2$ are distinct, an eigenbasis must consist two elements of $T_\ell (E_1)$ and two elements of $T_\ell(E_2)$, but there is no basis of $T_\ell(A)$, the lattice of elements of $T_\ell(E_1)+ T_\ell(E_2)$ where both parts are congruent mod $\ell$, consisting only of elements of $T_\ell(E_1)$ and $T_\ell(E_2)$. If the eigenvalues of Frobenius of $E_1$ and $E_2$ are equal, then they are isogenous.
Of course this is no problem as there are infinitely many such primes (https://arxiv.org/abs/1411.2914).
However, one can take a product of many such abelian surfaces for many different pairs of elliptic curves. The product is only semisimple if each pair of curves is isogenous. Standard heuristics suggest there should be only finitely many primes where all these pairs are isogenous (already 3 independent pairs should be sufficient).
